A trolley of height 0.2m moving on a horizontal bench of height 3.2m strikes a barrier at the edge of the bench. The object on top of the trolley flies off on impact and lands on the ground 2.5m from

Height of bench $= 3.2 \text{ m}$  
Height of trolley $= 0.2 \text{ m}$  
Total vertical height, $h = 3.2 \text{ m} + 0.2 \text{ m} = 3.4 \text{ m}$

Since the object flies off horizontally, its initial vertical velocity $u_y = 0 \text{ m/s}$.  
Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$.
$$
h = u_y t + \frac{1}{2}gt^2
$$
$$
3.4 \text{ m} = (0 \text{ m/s})t + \frac{1}{2}(9.8 \text{ m/s}^2)t^2
$$
$$
3.4 = 4.9t^2
$$

$$
t^2 = \frac{3.4}{4.9}
$$
$$
t = \sqrt{\frac{3.4}{4.9}}
$$
$$
t \approx 0.833 \text{ s}
$$
The time taken by the object to reach the ground is $\boxed{\text{0.833 s}}$.

Mass of trailer, $m_1 = 30 \text{ tonnes} = 30 \times 1000 \text{ kg} = 30000 \text{ kg}$  
Initial velocity of trailer, $u_1 = 72 \text{ km/h} = 72 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 20 \text{ m/s}$  
Mass of bus, $m_2 = 10 \text{ tonnes} = 10 \times 1000 \text{ kg} = 10000 \text{ kg}$  
Initial velocity of bus, $u_2 = 0 \text{ m/s}$ (stationary)

The total momentum before collision equals the total momentum after collision.
$$
m_1 u_1 + m_2 u_2 = (m_1 + m_2) V
$$
where $V$ is the common velocity after impact.

$$
(30000 \text{ kg})(20 \text{ m/s}) + (10000 \text{ kg})(0 \text{ m/s}) = (30000 \text{ kg} + 10000 \text{ kg}) V
$$
$$
600000 \text{ kg m/s} + 0 = (40000 \text{ kg}) V
$$
$$
600000 = 40000 V
$$
$$
V = \frac{600000}{40000}
$$
$$
V = 15 \text{ m/s}
$$
The common velocity is $\boxed{\text{15 m/s}}$.